For a polynomial $f=X^n+a_1X^{n-1}+\ldots+a_n \in \mathbb{Q}[X]$ we define $\varphi(f):=a_1 \in \mathbb{Q}$. Now I want to show that for the $n$th cyclotomic polynomial $\Phi_n$ it holds that $$\varphi(\Phi_n)=-\mu(n)$$
where $\mu(n)$ is the Möbius function. What I know is that $\displaystyle\Phi_n=\prod_{d|n} (X^{\frac{n}{d}}-1)^{\mu(d)}$.

Do you know the relation between $a_1$ and the roots of $f$?
–
Gerry MyersonApr 3 '14 at 8:46

1

To elaborate on @GerryMyerson's hint, recall that for a monic polynomial of degree $n$, the coefficient of the degree $n-1$ term is equal to the negative of the sum of the roots of the polynomial.
–
heropupApr 3 '14 at 8:57

Update: For monic univariate polynomials $f_1,f_2$ of degree at least $1$
we have
$$\left[f_1(X)\,f_2(X)\right]_{-1} = \left[f_1(X)\right]_{-1}+\left[f_2(X)\right]_{-1}$$
because the coefficient in question is the negative sum of the roots
of $f_1$ and $f_2$ (with multiplicity).
Even without considering roots,
this follows from looking at how the product expands.

Update: Proof
that $\mu$ is the unique arithmetic function with property $(1)$:

Now for uniqueness:
Let $\mu_1,\mu_2$ be arithmetic functions with property $(1)$.
Then necessarily $\sum_{d\mid n}\mu_1(d)=\sum_{d\mid n}\mu_2(d)$
for all positive integers $n$.
Suppose $\mu_1\neq \mu_2$, then there exists a minimal positive integer $m$
such that $\mu_1(m)\neq \mu_2(m)$. But then
$\sum_{d\mid m}\mu_1(d)\neq\sum_{d\mid m}\mu_2(d)$ which contradicts the
hypothesis. (All $d<m$ do not make any difference since $m$ is minimal
in that respect, but $d=m$ does make a difference.)
The only remaining possibility is $\mu_1=\mu_2$.